A greenhouse in a? tri-county area has kept track of its customers for the last several years and has determined that? 28% of its customers plant a vegetable garden in the spring. the greenhouse obtains a random sample of 1000 of its customers. what is the mean of the sampling distribution of modifyingabove p with caretp?, the sample proportion of customers that plant a vegetable garden in the? spring?

We are given two values
p = 28%
and
n = 1000
We are asked to get the mean of the sampling distribution of the customers that plant a vegetable in the spring.

We simply use the formula
p̂ = np
So, substituting
p̂ = 0.28 (1000)
p̂ = 280
The mean is 280

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joaobezerra joaobezerra · Answer 2 · 2019-12-02T13:26:13+01:00

Answer:

The mean is 280 customers.

The sample proportion is 28%.

Explanation:

For each customer, there are only two possible outcomes. Either they plant a vegetable in the spring, or they do not. This means that we can solve this problem using concepts of the binomial probability distribution.

Binomial probability distribution.

Probability of exactly x sucesses on n repeated trials, with p probability. Has an expected value of:

[tex]E(X) = np[/tex]

In this problem, we have that:

There are 1000 customers, so [tex]n = 1000[/tex].

28% of its customers plant a vegetable garden in the spring. This means that [tex]p = 0.28[/tex].

So:

What is the mean of the sampling distribution of modifyingabove p with caretp?

The mean is the expected value, so:

[tex]E(x) = np = 1000*0.28 = 280[/tex]

The mean is 280 customers.

The sample proportion of customers that plant a vegetable garden in the? spring?

It is the same as the population proportion, so the sample proportion is 28%.